Question

If $\frac{dy}{dx}=\frac{y^3}{e^{2x}+y^2}$ and $y\left(0\right)=1,$ then

• A. $y^2=e^{2x}-2e^{2x}\ln y$
• B. $y^2=e^{2x}+2e^{2x}\ln y$
• C. $y^2=e^{2x}+\frac{1}{2}{e}^{2x}\ln y$
• D. $y^2=e^{2x}-\frac{1}{2}{e}^{2x}\ln y$

Answer 1

The correct option is A $y^2=e^{2x}-2e^{2x}\ln y$

From given equation,
$\frac{dx}{dy}=\frac{e^{2x}+y^2}{y^3}$
$\Rightarrow \frac{dx}{dy}=\frac{1}{y}+e^{2x}\cdot \frac{1}{y^3}$
$\Rightarrow \frac{dx}{dy}-\frac{1}{y}=e^{2x}\cdot \frac{1}{y^3}$
$\Rightarrow e^{-2x}\cdot \frac{dx}{dy}-\frac{1}{y}\cdot e^{-2x}=\frac{1}{y^3}$

Put $e^{-2x}=t$
$\Rightarrow -2e^{-2x}\frac{dx}{dy}=\frac{dt}{dy}$
$\Rightarrow e^{-2x}\frac{dx}{dy}=-\frac{1}{2}\frac{dt}{dy}$
$\therefore -\frac{1}{2}\frac{dt}{dy}-\frac{1}{y}\times t=\frac{1}{y^3}$
$\Rightarrow \frac{dt}{dy}+\left(\frac{2}{y}\right)t=-\frac{2}{y^3}...\left(i\right)$

I.F. $=e^{2\int \frac{dy}{y}}=e^{2\ln y}=y^2$
$\therefore$ Solution of equation $\left(i\right)$ is
$t\times y^2=-2\int \frac{1}{y^3}\times y^{2}dy$
$\Rightarrow e^{-2x}\times y^2=-2\ln y+C$
When $x=0,y\left(0\right)=1$
$\therefore e^0\times 1=-2\ln 1+C$
$\Rightarrow C=1$
$\therefore e^{-2x}\cdot y^2=-2\ln y+1$
$\Rightarrow y^2=-2e^{2x}\ln y+e^{2x}$